What is the solution for y^'=2y+e^{2x}cos(3x),y(0)=4 ?

The solution for y^'=2y+e^{2x}cos(3x),y(0)=4 is y=(sin(3x))/(3e^{-2x)}+4e^{2x}

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y^'=2y+e^{2x}cos(3x),y(0)=4

y^{^{'}} = 2 y + e^{2 x} cos (3 x), y (0) = 4

y = \frac{sin ( 3 x )}{3 e ^{- 2 x}} + 4 e^{2 x}

Solution steps

y^{'} (x) = 2 y + e^{2 x} cos (3 x)

Solve linear ODE:

y = \frac{sin ( 3 x )}{3 e ^{- 2 x}} + 4 e^{2 x}

y = \frac{sin ( 3 x )}{3 e ^{- 2 x}} + 4 e^{2 x}

\frac{6 x ^{2} + 6 x + 6}{x}

\int sin (4 t) d t

\int sin (x) sin (2 x) d x

y^{^{'}} = 4 x + y, y (0) = 1

\int_{0}^{6} 6 x - (3 x - 12) d x

What is the solution for y^'=2y+e^{2x}cos(3x),y(0)=4 ?
The solution for y^'=2y+e^{2x}cos(3x),y(0)=4 is y=(sin(3x))/(3e^{-2x)}+4e^{2x}